How to integrate $\int\limits_{0}^{\pi/2}\frac{dx}{\cos^3{x}+\sin^3{x}}$? I have$$\int\limits_{0}^{\pi/2}\frac{\text{d}x}{\cos^3{x}+\sin^3{x}}$$
Tangent half-angle substitution gives a fourth-degree polynomial in the denominator that is difficult to factor.
 A: One may write
$$
\begin{align}
\int_0^{\pi/2}\frac{\text{d}x}{\cos^3{x}+\sin^3{x}}&=\int_0^{\pi/2}\frac{\text{d}x}{(\cos x+\sin x)(\cos^2{x}-\cos x\sin x+\sin^2{x})}
\\\\&=\frac{\sqrt{2}}2\int_0^{\pi/2}\frac{\text{d}x}{\cos(x-\frac{\pi}4)\:(1-\frac12\sin(2x))}
\\\\&=\frac{\sqrt{2}}2\int_{-\pi/4}^{\pi/4}\frac{\text{d}u}{\cos u\:\left(1-\frac12\cos(2u)\right)}
\\\\&=\sqrt{2}\int_0^{\pi/4}\frac{\cos u\:\text{d}u}{\left(1-\sin^2u\right)\:\left(\sin^2u+\frac12\right)} 
\\\\&=\sqrt{2}\int_0^{\sqrt{2}/2}\frac{\text{d}v}{\left(1-v^2\right)\:\left(v^2+\frac12\right)} 
\\\\&=\frac{\pi}3+\frac{2\sqrt{2}}{3}\log\left(1+\sqrt{2}\right).
\end{align}
$$
A: $$I = \int_{0}^{\pi/2}\frac{dx}{\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\left(1-\frac{1}{2}\sin(2x)\right)}=\int_{-\pi/4}^{\pi/4}\frac{dx}{\sqrt{2}\cos(x)\left(1-\frac{1}{2}\cos(2x)\right)} $$
hence, through the substitution $x=\arcsin t$:
$$ I = \sqrt{2}\int_{0}^{\pi/4}\frac{dx}{\cos(x)\left(\frac{3}{2}-\cos^2 x\right)}=\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\frac{dt}{(1-t^2)\left(\frac{1}{2}+t^2\right)}$$
and the last integral is perfectly manageable through partial fraction decomposition.
The outcome is:

$$ \int_{0}^{\pi/2}\frac{d\theta}{\sin^3\theta+\cos^3\theta} = \color{red}{\frac{\pi}{3}+\frac{2\sqrt{2}}{3}\,\log\left(1+\sqrt{2}\right)}.$$

